Mathematics Questions and Answers – Trigonometric Equations – 1

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This set of Mathematics Multiple Choice Questions & Answers (MCQs) focuses on “Trigonometric Equations – 1”.

1. The solutions of a trigonometric equation for which ___________ are called principal solutions.
a) 0 < x < 2π
b) 0 ≤ x < π
c) 0 ≤ x < 2π
d) 0 ≤ x < nπ
View Answer

Answer: c
Explanation: The solutions of a trigonometric equation for which 0 ≤ x < 2π are called principal solutions. x should lie between 0 and 2π except 2π.
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2. The expression involving integer ‘n’ which gives all solutions of a trigonometric equation is called the principal solution.
a) True
b) False
View Answer

Answer: b
Explanation: The expression involving integer ‘n’ which gives all solutions of a trigonometric equation is called the general solution. n is used in general solution only.

3. Find the principal solutions of the equation cos x = 1/2.
a) π/6, 5π/6
b) π/3, 5π/3
c) π/3, 2π/3
d) 2π/3, 5π/3
View Answer

Answer: b
Explanation: Since cos x is positive in 1st and 4th quadrant, so there are two principal solutions to above equation i.e. x = π/3, 2π-π/3. So, x = π/3, 5π/3.
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4. Find the principal solutions of the equation tan x = – √3.
a) π/6, 5π/6
b) π/3, 5π/3
c) π/3, 2π/3
d) 2π/3, 5π/3
View Answer

Answer: d
Explanation: Since tan x is negative in 2nd and 4th quadrant so, there are two principal solutions to above equation i.e. x = π – π/3, 2π – π/3. So, x=2π/3, 5π/3.

5. Find the principal solutions of the equation sec x = -2.
a) 2π/3, 4π/3
b) 2π/3, 5π/3
c) 4π/3, 5π/3
d) π/3, 5π/3
View Answer

Answer: a
Explanation: Since sec x is negative in 2nd and 3rd quadrant so, there are two principal solutions to above equation i.e. x = π – π/3, π + π/3. So, x=2π/3, 4π/3.
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6. Find the principal solutions of the equation sin x = -1/√2.
a) π/4, 3π/4
b) 3π/4, 5π/4
c) 3π/4, 7π/4
d) 5π/4, 7π/4
View Answer

Answer: d
Explanation: Since sin x is negative in 3rd and 4th quadrant so, there are two principal solutions to above equation i.e. x = 2π – π/4, 2π + π/4. So, x=5π/4, 7π/4.

7. Find the principal solutions of the equation cosec x=2.
a) π/6, 5π/6
b) π/6, 7π/6
c) 5π/6, 11π/6
d) 5π/6, 7π/6
View Answer

Answer: a
Explanation: Since cosec x is positive in 1st and 2nd quadrant, so there are two principal solutions to above equation i.e. π/6, π – π/6. So, x=π/6, 5π/6.
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8. Find the principal solutions of the equation cot x=1/√3.
a) π/3, 4π/3
b) 2π/3, 5π/3
c) 4π/3, 5π/3
d) π/3, 5π/3
View Answer

Answer: a
Explanation: Since cot x is positive in 1st and 3rd quadrant, so there are two principal solutions to above equation i.e. π/3, π + π/3. So, x=π/3, 4π/3.

9. Find general solution to equation sin x = 1/2.
a) x = nπ + (-1)n π/3
b) x = nπ + (-1)n π/6
c) x = nπ + (-1)n 2π/3
d) x = nπ + (-1)n 5π/6
View Answer

Answer: d
Explanation: sin x= 1/2
sin x = sin π/6
x = nπ + (-1)n π/6.
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10. Find general solution to equation cos x = – 1/√2.
a) x = 2nπ±7π/4
b) x = 2nπ±3π/4
c) x = 2nπ±π/4
d) x = 2nπ±π/3
View Answer

Answer: c
Explanation: cos x = – 1/√2
cos x = – cos π/4 = cos (π- π/4) = cos 3π/4
So, x = 2nπ±π/4.

11. Find general solution to equation cot x = √3.
a) x = nπ + π/3
b) x = nπ + π/6
c) x = nπ + 2π/3
d) x = nπ + 5π/6
View Answer

Answer: b
Explanation: cot x = √3
tan x = 1/√3
tan x = tan π/6
x = nπ + π/6.

12. Solve: tan x = cot x
a) x = nπ/2 + (π/4)
b) x = nπ + (3π/4)
c) x = nπ + (π/4)
d) x = nπ/2 + (3π/4)
View Answer

Answer: a
Explanation: tan x = cot x
tan x = cot (π/2 – x)
x = nπ + (π/2 – x)
2x = nπ + (π/2)
x = nπ/2 + (π/4).

Sanfoundry Global Education & Learning Series – Mathematics – Class 11.

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Manish Bhojasia, a technology veteran with 20+ years @ Cisco & Wipro, is Founder and CTO at Sanfoundry. He is Linux Kernel Developer & SAN Architect and is passionate about competency developments in these areas. He lives in Bangalore and delivers focused training sessions to IT professionals in Linux Kernel, Linux Debugging, Linux Device Drivers, Linux Networking, Linux Storage, Advanced C Programming, SAN Storage Technologies, SCSI Internals & Storage Protocols such as iSCSI & Fiber Channel. Stay connected with him @ LinkedIn | Youtube | Instagram | Facebook | Twitter